Question: Simplify; express your answer in exponential form. Assume $n\neq 0, r\neq 0$. $\dfrac{{(n^{-2})^{-4}}}{{(n^{-2}r)^{3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{-2}}$ to the exponent ${-4}$ . Now ${-2 \times -4 = 8}$ , so ${(n^{-2})^{-4} = n^{8}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-2}r)^{3} = (n^{-2})^{3}(r)^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{-2})^{-4}}}{{(n^{-2}r)^{3}}} = \dfrac{{n^{8}}}{{n^{-6}r^{3}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{8}}}{{n^{-6}r^{3}}} = \dfrac{{n^{8}}}{{n^{-6}}} \cdot \dfrac{{1}}{{r^{3}}} = n^{{8} - {(-6)}} \cdot r^{- {3}} = n^{14}r^{-3}$.